Organic Chemistry 1)Compound A (C22H16) decolorizes a solution of bromine in CCl

Question

Organic Chemistry

1)Compound A (C22H16) decolorizes a solution of bromine in CCl4.  Independent analysis showsthat compound A does not possess any phenyl rings. In the presence of Pt and excess H2 compound A undergoesconversion to compound B (C22H28). How many rings does compound A possess?

Question 5 [1.5 pts] Which one of the following statements is correct? one of the following statements is Al The IR absorption band due to bond C will be less intense than the IR absorption band due to bond D The IR absorption band due to bond C will be more intense than the IR absorption band due to bond D None of the above

Explanation / Answer

Answer – 1) Given compound A molecular formula C22H16 and it is decolorizes a solution of bromine in CCl4­, means the compound A is with double bond.

When compound A is reacted H2/Pt there is formed compound B with molecular formula C22H28.

Now we need to calculate the site of unsaturation in compound A

Unsaturation in compound A = C22H46 - C22H16

                                     = 30 H /2 H

                                     = 15

So there are 15 site of unsaturation and in compound B there are increased H from 16 to 28 means 12 H added by hydrogenation, so there are 6 double bond.

So remaining site of unsaturation = 15 -6 = 9

So there are 9 rings in the compound A.

2) In this one there is given bond D is C-Cl bond and in the IR it is showing the stretching frequency between 600-800 strong and the bond C is N-Al bond and its stretching frequency between the

We know the as the bond length increase means it is gets already stretched then it need very low energy for the stretching the bond between two atom and it show at lower frequency value and those bond are strong like between the strong electronegative and electropositive atoms like H-Cl, H-F O-H show at the higher IR shift value.

We know the electronegative difference between the Cl and C is 0.61 and between the N and Al is 1.43, so N-Al bond is more polar than C-Cl

So, N-Al needed more energy than C-Cl bond, so it is showing the more intense than C-Cl.

So, the IR absorption band due to the bond C will be more intense than the IR absorption band due to bond D.

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